The binding energy of deuteron {}_1^2H is 1.1 | vedclass

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(C) The binding energy per nucleon for deuteron $({}_1^2H)$ is $1.112 \, MeV$. Since it has $2$ nucleons,the total binding energy of one deuteron is $

Vedclass · Accepted Answer

$23.8$

Vedclass · Answer

(C) The binding energy per nucleon for deuteron $({}_1^2H)$ is $1.112 \, MeV$. Since it has $2$ nucleons,the total binding energy of one deuteron is $2 	imes 1.112 \, MeV = 2.224 \, MeV$.In the reaction ${}_1^2H + {}_1^2H 	o {}_2^4He + Q$,the initial total binding energy is $2 	imes (2.224 \, MeV) = 4.448 \, MeV$.The $\alpha$-particle $({}_2^4He)$ has $4$ nucleons,and its binding energy per nucleon is $7.047 \, MeV$. Thus,the total binding energy of the final product is $4 	imes 7.047 \, MeV = 28.188 \, MeV$.The energy released $Q$ is the difference between the total binding energy of the products and the reactants: $Q = [B.E.]_{final} - [B.E.]_{initial}$.$Q = 28.188 \, MeV - 4.448 \, MeV = 23.74 \, MeV$.Rounding to the nearest provided option,$Q \approx 23.8 \, MeV$.

The binding energy of deuteron ${}_1^2H$ is $1.112 \, MeV$ per nucleon and an $\alpha$-particle ${}_2^4He$ has a binding energy of $7.047 \, MeV$ per nucleon. Then in the fusion reaction ${}_1^2H + {}_1^2H \to {}_2^4He + Q$,the energy $Q$ released is ......... $MeV$.

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